The concept of continuous function is basic to much of mathematics.
Continuous functions on the real line appear in the first pages of any
calculus book, and continuous functions in the plane and in space follow
not far behind. More general kinds of continuous functions arise as one
goes further in mathematics. In this section, we shall formulate a
definition of continuity that will include all these as special cases,
and we shall study various properties of continuous functions. Many of
these properties are direct generalizations of things you learned about
continuous functions in calculus and analysis.
Continuity of a Function
Let and be a topological spaces. A function
is said to be
continuous if for each open subset of , the set is an open subset of .
Recall that is the set
of all points of for which ; it is empty if does not intersect the image set of .
Continuity of a function depends not only upon the function itself, but also on the topologies
specified for its domain and range. If we wish to emphasize this fact,
we can say that is
continuous relative to specific topologies on
and .
Let us note that if the topology of the range space is given by a basis , then to prove continuity of
it suffices to show that the
inverse image of every basis element is open. Then arbitrary open set
of can be written as a union of basis
elements Then so that is open
if each set is
open.
If the topology on is given by
a subbasis , to prove
continuity of it will even
suffice to show that the inverse image of each subbasis element is open.
The arbitrary basis element for
can be written as a finite
intersection of subbasis element; it follows from the equation that the inverse image of every basis element is open.
EXAMPLE 1 Let us consider a function like those
studied in analysis, a "real-valued function of a real variable," In analysis, one defines continuity of f via the " definition". As one
would expect, the
definition and ours are equivalent To prove
that our definition implies the definition, for instance, we proceed as follows.
Given in , and given , the interval is an open set of the range space . Therefore, is an open set
in the domain space .
Because contains the
point , it contains some basis
element about . We choose to be the smaller of the two
numbers and . Then if , the point must be in , so that , and , as
desired.
Conversely, given in , and given , , if , we have . Let , then , thus is a open set, because is a basis
element, as desired.
EXAMPLE 2 In calculus one considers the property of
continuity for many kinds of functions. For example, one sudies
functions of the following types: Each of them has a notion of continuity defined for it. Our
general definition of continuity includes all these as special cases,
this fact will be a consequence of general theorems we shall prove
concerning continuous functions on product spaces and on metric
spaces.
EXAMPLE 3 Let denote the set of real numbers
in its usual topology, and let denote the same set in the
lower limit topology. Let be the indentity function. for every real number . Then is not a continuous function, the
inverse image of the open set
of equals itself,
which is not open in . On
the other hand, the identity function is continuous, because the inverse image of is itself, which is open in .
Theorem 18.1. Let and be topological spaces; let . Then the following are
equivalent:
is continuous.
For every subset of , one has .
For every closed set of
, the set is closed in .
For each and each
neighborhood of , there is a neighborhood of such that .
If the condition in (4) holds for the point of , we say that is continuous at the point
.
proof. We show
that (1) (2) (3) (1) and that (1) (4) (1).
(1) (2). Assume that
is continuous. Let be a subset of . We show that if , then . Let be a neighborhood of . Then is an open set of containing ; it must intersect in some point . THen intersects in the point , so that , as
desired.
(2) (3). Let be closed in and let . We wish to prove that is closed in ; we show that . By elementary set theory, we
have .
Therefore, if , so that . Thus , so that . Thus , so that , as desired.
(3) (1). Let be an open set of . Set . Then Now is a closed set of
. Then is closed in by hypothesis, so that is open in , as desired.
(4) (1). Let be an open set of ; let be a point of . Then , so that by hypothesis there
is a neighborhood of such that . Then . If follows that
can be written as the
union of the open sets , so that
it is open.
(1) (4). Let and let be a neighborhood of . Then the set is a neighborhood of such that .
Homeomorphisms
Let and be topological spaces; let be a bijection. If both the
function and the inverse function
are continuous, then is
called a homeomorphism.
The condition that be
continuous says that for each open set of , the inverse image of under the map is open in . But the inverse image of under the map is the same as the image of under the map . See Figure 18.1. So another way to
define a homeomorphism is to say that it is a bijective correspondence
such that is open if and only if is open.
面对有关矩阵的题目，我们也要注意将矩阵看作一个线性变换，例如一个可逆矩阵可看作一个同构映射；一个上三角矩阵，可看作不变子空间分解；一个分块对角矩阵，可看作线性空间的直和分解，有根子空间直和分解，循环子空间直和分解，还有不变子空间直和分解，三者分别对应Jordan
form、frobenius form 和 直接对角化。
A subset of a topological
space is said to be
closed if the set is open.
EXAMPLE 1 The subset of is closed because its
complement is open. Similarly, is closed, because its
complement is open.
These facts justify our use of the terms "closed interval" and "closed
ray" The subset of is neither open nor
closed.
EXAMPLE 2 In the plane , the set is closed, because its complement is the union of the two sets
each of which is a product of open sets of and is, therefore, open in
EXAMPLE 3 In the finite complement
topology on a set , the
closed sets consist of itself and
all finite subsets of
finite complement topology
The finite complement topology (or cofinite topology) on a set is defined by specifying the open sets
in the topology:
1.The empty set and
the set itself are open.
2.Any subset is
open if and only if its complement is finite.
This topology is particularly useful when studying properties of
convergence and compactness, as it ensures that every sequence in has a convergent subsequence.
EXAMPLE 4 In the discrete topology on the set , every set is open, it
follows that every set is closed as well.
EXAMPLE 5 Consider the following subset of the real
line: in the subspace topology. In this space, the set is open, since it
is the intersection of the open set of with . Similarly, is open as a subset of ; it is even open as a subset of . Since and are complements in of each other, we conclude that both
and are closed as subsets of .
Let's talk about the properties of the closed sets.
Theorem 17.1. Let be a topological space. Then the
following condition hold:
and are closed.
Arbitrary intersections of closed sets are closed.
Finite unions of closed sets are closed.
proof. (1) and are closed because they are the
complements of the open sets and
, respectively.
(2) Given a collection of closed sets , we apply DeMorgan's law, Since the sets are open by definition, the right side of this
equation represents an arbitrary union of open sets, and is thus open.
Therefore, is
closed.
(3) Similarly, if is closed for
, consider the
equation The set on the right side of this equation is a finite
intersection of open sets and is therefore open, Hence is closed.
We should notice the conceptions of "closed set" is relative.
Now when dealing with subspaces, one needs to be careful in using the
term "closed set." If is a
subspace of , we say that a set
is closed in if is a subset of and if is closed in the subspace
topology of (that is, if
— is open in )
We have the following theorem:
Theorem 17.2. Let be a subspace of . Then a set is closed in if and only if it equals the
intersection of a closed set of
with .
proof. Assume
that , where is closed in . (See Figure 17.1) Then is open in , so that is open in , by the definition of the subspace
topology. But .
Hence is open in , so that is closed in . Conversely, assume that is closed in . (See Figure 17.1) Then is open in , so that by definition it equals the
intersection of an open set of
with . The set is closed in , and , so that
equals the intersection of a closed set of with , as desired.
Figure 17.1
Theorem 17.3. Let be a subspace of . If is closed in and is closed in then is closed in .
proof. If is closed in , then equals the intersection of a closed set
of with , by the theorem 17.2, so we have . And is also closed in , thus is closed in .
Closure and Interior of a
Set
Given a subset of a
topological space , the
interior of is
defined as the union of all open sets contained in .
The closure of is defined as the intersection of all
closed sets containing
The interior of is denoted by
Int and the closure of is denoted by . Obviously Int is an open set and is a closed set; furthermore,
If is open , ; while if is closed, .
We shall not make much use of the intenor of a set, but the closure
of a set will be quite important.
NOTE:
When dealing with a topological space and a subspace , one needs to exercise care in taking
closures of sets If is a subset
of , the closure of in and the closure of in will in general be different In such a
situation.
We reserve the notation to
stand for the closure of in . The closure of in can be expressed in terms of .
Theorem 17.4. Let be a subspace of , let be a subset of , let denote the closure of in . Then the closure of in equals .
proof. Let denote the closure of in . The set is closed in , so is closed in in
Theorem 17.2. Since
contains , and since by definition
equals the intersection of all
closed subsets of containing
, we must have
On the other hand, we know that
is closed in . Hence by Theorem
17.2, for some set
closed in . Then is a closed set containing ; because is the intersection of all such
closed sets, we conclude that . Then .
Since the collection of all closed sets in , like the collection of all open sets,
is usually much too big to work with. Another way of
describing the closure of a set, useful because it involves only
a basis for the topology of , is given in the following
theorem.
Theorem 17.5. Let be a subset of the topological space
.
Then if and
only if every open set containing
intersects .
Supposing the topology of
is given by a basis, then if and only if every basis element containing intersects .
proof. Consider
the statement in (a). It is a statement of the form . Let us transform
each implication to its contrapositive, thereby obtaining the logically
equivalent statement (not ) (not ). Written out, it is the following.
In this form, our theorem is easy to prove. If is not in , the set is an open set containing
that does not intersect , as desired. Conversely, if there
exists an open set containing
which does not intersect , then is a closed set containing . By definition of the closure , the set must contain , therefore, cannot be in . Statement (b) follows readily if
every open set containing
intersects , so does every basis
element containing , because is an open set. Conversely, if every
basis element containing
intersects , so does every open
set containing , bacause contains a basis element that contains
.
Mathematicians often use some special terminology here. They shorten
the statement " is an open set
containing " to the phrase
Using this terminology, one can write the first
half of the preceding theorem as follows:
If is a subset of the
topological space , then if and only if every
neighborhood of intersects .
EXAMPLE 6 Let
be the real line . If
, then , for every neighbrhood of
intersects , while every point outside has a neighborhood disjoint from
Similar arguments apply to the
following subsets of .
If , then . If , then . If is
the set of rational numbers, then . If is
the set of positive integers, then . If
is the set of positive
reals, then the closure of is the set .
EXAMPLE 7 Consider the subspace of the real line . The set is subset of , its closure in is the set , and its closure in
is the set
Limit Point
There is yet another way of describing the closure of a set, a way
that involves the important concept of limit point, which we consider
now.
If is a subset of the
topological space and if is a point of , we say that is a limit
point (or "cluster point," or "point of accumulation") of
if every neighborhood of
intersects in some point other than itself. Said differently,
is a limit point of if it belongs to the closure of
. The point
may lie in or not; for this definition it does not
matter.
EXAMPLE 8 Consider the real line . If , then the point is a limit point of and so is the point . In fact, every point of the
interval is a limit point of
, but no other point of is a limit point of .
If , then is
the only limit point of . Every
other point of has a neighborhood that either
does not intersect at all, or it
intersects only in the point
itself. If , then the limit
points of are the points of the
interval .
If is the set of
rational numbers, every point of is a limit point of . If is the set of positive
integers, no point of is
a limit point of . If
is the set of positive
reals, then every point of is a limit point of .
Theorem 17.6. Let be a subset of the topological space
, let be the set of all limit points of
. Then
proof. If in , every neighborhood of intersects (in a point different from x).
Therefore, by Theorem 17.5,
belong to . Hence . Since by
definition , it
follows that .
To demonstrate the reverse inclusion, we let be a point of and show that . If happens to lie in , it is trivial that ; suppose that does not lie in . Since , we know that every neighborhood of intersects ; because , the set must
intersect in a point different
from . Then , so that , as desired.
Corollary 17.7 A subset of a topological space is
closed if and only if it contains all its limit points.
proof. The set
is closed , and the latter holds
.
Hausdorff Spaces
One's expenence with open and closed sets and limit points in the
real line and the plane can be misleading when one considers more
general topological spaces.
For example, in the spaces and , each one-point
set is closed.
This fact is easily proved; every point different from has a neighborhood not intersecting
, so that is its own closure.
But this fact is not true for arbitrary topological
spaces. Consider the topology on the three-point set indicated in Figure 17.3. In
this space, the one-point set is not closed, for its complement () is not open.
Figure 17.3
And, in topological space , a
subset of can have many different limit
points.
Similarly, one's experience with the properties of convergent
sequences in and can be misleading when one
deals with more general topological spaces. In an arbitrary topological
space, one says that a sequence of points of the space to the point of provided that, corresponding to each
neighborhood of , there is a positive integer such that for all . In and , a sequence cannot
converge to more than one point, but in an arbitrary space, it
can. In the space indicated in Figure 17.3, for example, the
sequence defined by setting
for all converges not only to the
point , but also to the
point and to the point !
Topologies in which one-point sets are not closed, or in which
sequences can converge to more than one point, are considered by
many mathematicians to be somewhat strange. They are not really
very interesting, for they seldom occur in other branches of
mathematics.
Therefore, one often imposes an additional condition that
will rule out examples like this one, bringing the class of spaces under
consideration closer to those to which one's geometric intuition
applies. The condition was suggested by the mathematician
Felix Hausdorff, so mathematicians have come to call it
by his name.
Definition. A topological space is called a Hausdorff
space if for each pair of distinct points of , there exists neighborhoods , and of and , respectively, that are disjoint.
Theorem 17.8. Every finite point set in a Hausdorff
space is closed.
proof. It
suffices to show that every one-point set is closed. If is
a point of different from , then and have disjoint neighborhoods and , respectively. Since does not intersect , the point cannot belong to the closure of set
. As a result, the closure
of the set is itself, so that is is
closed.
The condition that finite point sets be closed is in fact
weaker than the Hausdorff condition.
For example, the real line in the finite
complement topology is note a Hausdorff space, but it is a
space in which finite point sets are closed.
The condition that finite point sets be closed has been given a name
of its own: it is called the axiom
Theorem 17.9. Let be a space satisfying the axiom; let be a subset of . Then the point is a limit point of if and only if every neighborhood of
contains infinitely many points
of .
proof. If every
neighborhood of intersects in infinitely many points, it certainly
intersects in some point other
than itself, so that is a limit point of .
Conversely, suppose is a limit
point of , and suppose some
neighborhood of intersects in only finitely many points. Then
also intersects in finitely many points; let
be the
points of . The
set is
an open set of , since the finite
point set is
closed; then is a neighborhood of x that intersects the set not at all. This contradicts
the assumption that is a limit
point of .
One reason for our lack of interest in the axiom is the fact that many
of the interesting theorems of topology require not just that axiom, but
the full strength of the Hausdorff axiom. Furthermore, most of
the spaces that are important to mathematicians are Hausdorif spaces.
The following two theorems give some substance to these remarks.
Theorem 17.10. If is a Hausdorff space, then a sequence
of points of converges to at most
one point of .
proof. Suppose
that is a sequence of points of
that converges to . If , let and be disjoint neighborhoods of and , respectively. Since contains for all but finitely many values of
, the set cannot. Therefore, cannot converge to .
If the sequence of points of
the Hausdorff space converges to
the point of , we often write , and we say that is the limit
of the sequnce .
Theorem 17.11.
(1). Every simply ordered set is a Hausdorff space in the order
topology.
(2). The product of two Hausdor-if spaces is a Hausdorff space.
(3). A subspace of a Hausdorff space is a Hausdorff space.
proof.
(1). Given two elements of
the simply orderd set , and . We can find an interval
be the neighborhood of , and be the neighborhood of such that . Thus,
, is a Hausdorff space.
(2). Suppose that and are the two Hausdorff spaces, and is the product topology. Given
, then
. We can find a neighborhood of , and of such that , we also can
find a neighborhood of , and of such that . Hence , and . So is a Hausdorff space.
(3). Suppose is a Hausdorff
space, . Obviously,
is also a Hausdorff space.
If is a simply ordered set,
there is a standard topology for ,
defined using the order relation. It is called the order
topology; in this section, we consider it and study some
of its properties.
Suppose that is set having
simple order relation . Given
elements and of such that , there are four subsets of that are called the intervals
determined by and . They are the following:
The notation used here is familiar to you already in the case where X
is the real line, but these are intervals in
an arbitrary ordered set. A set of the first type is called an
open interval in X, a set of the last type is
called a closed interval in X, and sets of the
second and third types are called half-open
intervals
Definition. Let be a set with a simple order relation;
assume has more than one element.
Let be the collection
of all sets of the following types:
All open intervals in
.
All intervals of the form , where is the
smallest element (if any) of .
All intervals of the form , where is the
largest element (if any) of .
The collection is a
basis for a topology on , which is
called the order topology.
if has no smallest element,
there are no sets of type (2), and if X has no largest element, there
are no sets of type (3)
Note: The elements of the order topology are not the
intervals, we should notice the definition of the
interval!!!
We need to check that satisfies the requirements
for a basis.
First, note that every element
of lies in at least one element
of : The smallest
element (if any) lies in all sets of type (2), the largest element (if
any) lies in all sets of type (3), and every other element lies in a set
of type (1).
Second, note that the intersection of any two sets of the preceding
types is again a set of one of these types, or is empty.
EXAMPLE 1 The standard topology on , as defined in the preceding
section is just the order topology derived from the usual order on
EXAMPLE 2 Consider the set in the
dictionary order, we shall denote the general element of by , to avoid difficulty with
notation. The set has neither a largest nor a smallest
element, so the order topology on , so the order topology on has as basis
the collection of all open intervals of the form for , and for , and . These two types of intervals are
indicated below.The subcollection consisting of only intervals of the
second type is also a basis for the order topology on , it is easy
to check, just think a moment.
EXAMPLE 3 The positive integers form an orederd set with a
smallest element. The order topology on is discrete topology, for
every one-point set is open If , then the one-point set is a basis element;
and if , the one-point set is a basis element.
EXAMPLE 4 The set in the dictionary order is another
example of an ordered set with a smallest element. Denoting by and by , we can represent by The order topology on
is not the discrete topology. Most one-point sets are open, but there is
an exception --- the one-point set .Any open set containing b1 must contain a basis element
about (by definition), and any
basis element containing
contains points of the
sequence.
Rays
Definition. If is an ordered set, and is an element of , there are four subsets of that are called the
rays determined by They are the following: Sets of the first two types are called open
rays, and sets of the last two types are called
closed rays
The use of the term "open" suggests that open rays in X are open sets
in the order topology. And so they are.
Consider, for example, the ray . If has a
largest element , then equals the basis element
. If has no largest element, then equals the union of all basis
elements of the form , for
. In either case, is open. A similar argument
applies to the ray .
The open rays, in fact, form a subbasis for the order topology on
, as we now show.
Because the open rays are open in the order topology, the topology
they generate is contained in the order topology.
On the other hand, every basis element for the order topology equals a
finite intersection of open rays; the interval equals the intersection of and , while and , if they exist, are themselves
open rays. Hence the topology generated by the open rays contains the
order topology.
The Product Topology on X × Y
If and are topological spaces, there is a
standard way of defining a topology on the cartesian product .
Definition. Let and be topological spaces. The
product topology on is the topology having as a
basis the collection of
all sets of the form ,
where is an open subset of and is an open subset of .
Let's check that is
a basis.
The first condition is trivial, since is itself a basis element. The second condition is
almost as easy, since the intersection of any two basis elements and is another basis element.
For and the latter set is a basis element because and are open in and , respectively. See the figure below, it
can help you understand it intutively.
Note: The collection is not a topology on . The union of the two
rectangles pictured in the figure above, for instance, is not a product
of two sets, so it is cannot belong to ; however, it is open in .
We may ask what's the basis of the topology of
Theorem 15.1. If is a basis for the topology
of and is a basis for the topology
of , then the collection is a basis for the topology of .
proof. We apply
Lemma 13.2. Given an open set of
and a point of , by definition of the product topology
there is a basis element
such that . Because and
are bases for and , respectively, we can choose an element
such that , and an element such that . Then . Thus
the collection meets
the criterion of Lemma 13.2, so is a basis for .
EXAMPLE 1 We have a standard topology on the order topology. The
product of this topology with itself is called the standard
topology on . It has as basis the collection of
all products of open sets of , but the theorem just proved
tells us that the much smaller collection of all products of open intervals in
will also serve as a
basis for the topology of . Each such set can be
pictured as the interior of a rectangle in
It is sometimes useful to express the product topology in
terms of a subbasis. To do this, we first define certain
functions called projections.
Definition. Let be defined by the equation let
be defined by the equation The maps and are called the projections of onto its first and second
factors, respectively.
We use word "onto" because
and are surjective.
If is an open subset of , then the set is precisely the set , which is open in . Similarly, if is open in , then , which is
also open in . The
intersection of these two sets is the set , as indicated in Figure below.
This fact leads to the following theorem:
Theorem 15.2. The collection is a subbasis for the product topology on .
proof. Let denote the product topology
on . let be the topology
generated by . Because
every element of
belongs to , so do the
arbitrary unions of finite intersections of elements of . Thus . On
the other hand, every basis element for the topology of is a finite intersection of
elements of , since
Therefore, , so that ,
hence , as desired.
The subspace Topology
Definition. Let be a topological space with topology
. If is a subset of , the collection is a topology on ,
called the subspace topology. With this
topology, is called a
subspace of ; its open sets consists of all
intersections of open sets of
with .
It is easy to see that is a topology.
It contains and because where and are elements of .
The fact that it is closed under finite intersections and arbitrary
unions follows from the equations
Lemma 16.1. If is a basis for the topology
of then the collection is a basis for the subspace topology on .
proof. Given
open in and given , we can choose an element such that . Then . It
follows from Lemma 13.2 that is a basis for the subspace
topology on .
When dealing with a space and
a subspace , one needs to be
careful when one uses the term "open set". Does one mean an
element of the topology of or an
element of the topology of ?
We make the following definition If is a subspace of , we say that a set U is open in
(or open relative to
) if it belongs to the topology of
; this implies in particular that
it is a subset of . We say that
U is open in if
it belongs to the topology of
There is a special situation in which every set open in is also open in .
Lemma 16.2. Let be a subspace of . If is open in and is open in , then is open in .
proof. Since
is open in , for some set open
. Since and are both open in , so is .
Now let us explore the relation between the subspace topology and the
order and product topologies.
Theorem 16.3. If is a subspace of and is a subspace of , then the product topology on is the same as the topology
inherits as a subspace of
.
proof.The set
is the general basis
element for , where is open in and is open in . Therefore, is the
general basis element for the subspace topology on . Now Since and are the general open sets fot
the subspace topologies on the
and , respectively, the set is the general
basis element for the product topology on .
Thus, we know that the bases for the subspace topology on and for the product topology on
are the same. Hence the
topologies are the same.
Now let be an ordered set in
the order topology, and let be a
subset of . The order relation on
, when restricted to , makes into an ordered set.
However, the resulting order topology on need not be the same as the topology
that inherits as a subspace of
. We give one example
where the subspace and order topologies on agree, and two examples where they do
not.
EXAMPLE 1 Consider the subset of the real line , in the subspace topology. The
subspace topology has as basis all sets of the form , where in an open interval in . Such a set is of one the
following types. By definition, each of these sets is open in . But sets of the second and third types
are not open in the larger space .
Note that these sets form a basis for the order topology on
Y. Thus, we see that in the case of the set , its subspace topology
(as a subspace of ) and
its order topology are the same.
EXAMPLE 2 Let
be the subset of
. In the subspace
topology on the the one-point set
is open,
because it is the intersection of the open set with . But in the order topology on , the set is not open. Any basis element for the
order topology on that contains
is of the form for some , such a
set necessarily contains points of less than .
EXAMPLE 3 Let . The dictionary order on is just the restriction to of the dictionary order on the
plane .
However, the dictionary order topology on is not the same as the subspace
topology on obtained
from the dictionary order topology on ..
For example, the set is open in in the subspace topology, but
not in the order topology, because we choose , any basis element for the order topology on
that contains is of the form
such a set is necessarily not be contained in
The set in the
dictionary order topology will be called the order
square, and denoted by
Convex
Given an ordered set , let us
say that a subset of is convex in
if for each pair of points of , the entire interval (a, b) of points
of lies in . Note that intervals and rays in are convex in .
Theorem 16.4 Let be an orderd set in the order topology;
let be a subset of that is convex in . Then the order topology on is the same as the topology inherits as a subspace of .
proof. Consider
the ray in . What is its intersection with ? If , then this is an open ray of the orderd set . If , then is either
a lower boud on or an upper bound
on , since is convex. In the former case, the set
equals all of
; in the latter case, is is
empty.
A similar work remark show that the intersection of the ray with is either an open ray of , or itself, or empty. Since the sets and form a subbasis for
the subspace topology on , and
since each is open in the order topology, the order topology contains
the subspace topology.
To prove the reverse, note that any open ray of Y equals the
intersection of an open ray of
with , so it is open in the
subspace topology on . Since
the open rays of are a
subbasis for the order topology on , this topology is contained in
the subspace topology, hence they are the same.
Comments: In this proof, we need to be familiar with
the definition of rays and its properties (such that it
is the subbasis for the order topology)
To avoid ambiguity, let us agree that whenever is an ordered set in the order topology
and is a subset of , we shall assume that is given the subspace topology unless
we specifically state otherwise. If is convex in , this is the same as the order topology
on , otherwise, it may not be.
Definition. If is any subset of , the exterior measure of
is where the infimum is taken over all countable coverings by
closed cubes. In general, the exterior measure is always non-negative
butt could be infinite, hence we have
Info Note
It is important to note that it would not suffice to allow finite
sumsin the definition of
One can, however, replace the coverings by cubes, with covering
by rectangles; or with coverings by balls. That the former alternative
yields the same exterior measure is quite direct.
EXAMPLE 1. The exterior measure of a point is zero.
The exterior measure of the empty set is also zero.
This is clear that a point is a cube with volume zero, and which
covers itself, so does the empty set.
EXAMPLE 2. The exterior measure of a closed cube is
equal to its volume.
Suppose is a closed cube in
. Since covers itself, we have .Therefore, it suffices to
prove the reverse inequality.
We consider an abitraty covering by cubes, and by the definition of
the exterior measure, it suffices to prove that For a fixed we choose for each an
open cube which contains , and such that . From the
open covering of the compact set , we may select a finite subcovering,
which after possibly renumbering the rectangles, we may write as . Taking
the closure of the cubes , we
have . Consequently, Since is
arbitrary, we find the inequality before we need to prove holds; thus
, as desired.
EXAMPLE 3. If
is an open cube, the result still holds.
Since is covered by its
closure , and , we immediately see
that , and obviously
by the definition
of exterior measure.
EXAMPLE 4. The exterior measure of a rectangle is equal to its volume.
Arguing as in Example 2, we see that . To obtain the reverse inequality, consider a grid
in formed by cubes of
sides length . Then, if
consists of the
(finite) collection of all cubes entirely contained in , and the (finite) collection
of all cubes that intersect the complement of , we first note that . Also, a simple argument yields Moreover, there are cubes in , and these cubes have
volume , so that . Hence and letting tend to
infinity yields , as
desired.
EXAMPLE 5. The exterior measure of is infinite.
This follows from the fact that any covering of is also a covering of any
cube , hence
. Since
can have arbitrarily large
volume, we must have
EXAMPLE 6. The Cantor set has exterior measure .
From the construction of , we know that , where
each is a disjoint
union of closed intervals,
each of length .
Consequently, for all , hence .
Observation 1 (Monotonicity) If , then .
This follows that any coverings of by a countable collection of cubes is
also a covering of .
In particular, monotonicity implies that every bounded subset of
has finite exterior
measure.
Observation 2 (Countable sub-additivity) If , then
.
First, we may assume that each , for otherwise the inequality clearly holds. For
any , the definition
of exterior measure yields for each a covering
by closed cubes with Then, is a covering of by closed cubes, and therefore Since this holds true for every , the second observation is
proved.
Observation 3 If , then , where the infimum is taken over all open
sets containing .
By monotonicity, it is clear that the inequality holds. For the
reverse inequality, let and choose cubes
such that , with Let denote an
open cube containing , and such
taht . Then is open, and by Observation 2
Hence , as was shown.
Observation 4 If , and , then
By Observation 2, we already know that , so it is
suffices to prove the reverse inequality. To this end, we first select
such that . Next, we
choose a covering by closed cubes, with . We may, after subdividing the cubes , assume that each has a diameter less than . In this case, each can intersect at most one of the two
sets or . If we denote by and the sets of those indices for which intersects and , respectively, then is empty, and we have Therefore
Observation 5 If a set is the countable union of
almost disjoint cubes, then
Let denote a cube
strictly contained in such that
, where is a arbitrary but fixed. Then,
for every , the cubes are disjoint, hence at a finite distance from one
another, and repeated applications of Observation 4 imply Since , we conclude that for every integer , When the tend to
infinity, we deduce for every , hence , as desired.
Info Note
This last property shows that if a set can be decomposed into almost
disjoint cubes, its exterior measure equals the sum of the volumes of
the cubes. Moreover, this also yields a proof that the sum is
independent of the decomposition
Measureable sets and
the Lebesgue measure
Definition. A subset of is Lebesgue
measurable or simply measureable, if for any
there exists an
open set with and
If is measurable, we define
its Lebesgue measure (or measure)
by Clearly, the Lebesgue measure inherits all the features
contained in Observations 1 - 5 of the exterior measure.
Let's study the properties of the Lebesgue measure.
Property 1 Every open set in is measurable.
Property 2 If , then is measurable.
In particular, if is a subset of
a set of exterior measure , then
is measurable.
By Observation 3 of the exterior measure, for every there exists an open set
with and . Since
, monotonicity implies , as
desired.
Property 3 A countable union of measurable sets is
measurable.
Suppose , where each is
measurable. Given ,
we may choose for each an open
set with and . Then the union is open, , nd , so monotonicity and sub-additivity of
the exterior measure imply
Property 4Closed sets are
measurable
First, we observe that it suffices to prove that compact sets
are measurable. Indeed, any closed set can be written as the union of compact
sets, say , where denotes the
closed ball of radius centered at
the origin; then Property 3 applies.
So suppose is compact (so that in
particular ), and
let . Since is
closed, the difference is open, and by Theorem 1.4 we may write this difference as a
countable union of almost disjoint cubes FOr a fixed , the finite
union is
compact; therefore
(we isolate this little fact in a lemma below). Since ,
Observation 1, 4, and 5 of the exterior measure imply Hence , and
this also holds in the limit as
tends to infinity. Invoking the sub-additivity property of the exterior
measure finally yields as desired.
Theorem 1.4. Every open subset of , , can be written as a countable
union of almost disjoint closed cubes.
Lemma 3.1. If
is closed, is compact, and these
sets are disjoint, then .
proof. Since
is closed, for each point , there exists so that . Since
covers , and is compact, we may find a subcover,
which we denote by . If
we let , then we must have . Indeed, if
and , then for some we have , and by
construction . Therefore and the lemma is proved.
Compact set: A bounded set is
compact if it is also closed. Compact sets
enjoy the Heine-Borel covering property.
Property 5 The complement of a measurable set is
measurable.
If E is measurable, then for every positive integer we may choose an open set with and .
The complement is
closed, hence is measurable, which implies that the union is also measurable by Property 3. Now we
simply note that ,
and such that for all .
Therefore, , and
is measurable by Property
2. Therefore is measurable
since it is the union of two measurable sets, namely and .
Property 6 A countable intersection of measurable
sets is measurable.
This follows from Properties 3 and 5, since
Warning
we have proved that the collection of measurable sets is closed under
countable unions and intersections. We emphasize, however, that the
operations of uncountable unions or intersections are
not permissible when dealing with measurable sets!
Theorem 3.2. If are disjoint measurable sets, and , then
proof. First, we
assume further that each is
bounded. Then, for each , by
applying the definition of measurability to , we can choose a closed subset
or with . For each fixed , the set are compact and
disjoint, so that . Since , we must
have Letting tend to
infinity, since was
arbitrary we find that Since the reverse inequality always holds(by sub-aditivity in
Observation 2), this concludes the proof when each is bounded.
In the general case, we select any sequence of cubes that increases
to , in the sense
that for all
and . We then let and for . If we
define measurable sets by . then The union above is disjoint and every is bounded. Moreover , and
this union is also disjoint. Putting these facts together, and using
what has already been proved, we obtain as claimed.
We make two definitions to state succinctly some further
consequences.
If is a countable collection of subsets of that increases to in the sense that for all , and , then we write
Similarly, if
decreases to in the sense that
for all k,
and ,
we write .
Corollary 3.3 Suppose are measurable subsets
of .
If , then
.
If and for some k, then
proof. For the
first part, let , , and in general for . By their construction, the sets
are measurable, disjoint, and
.
Hence and since we get the desired limit.
>For the second part, we may clearly assume that . Let for each , so that
is a disjoint union of measurable sets. As a result, we find that Hence, since , we see that , and the proof is
complete.
Info Note
The reader should note that the second conclusion may fail without
the assumption that for some . This is shown by the simple example
when , for all .
What follows provides an important geometric and analytic
insight into the nature of measurable sets, in terms of their
relation to open and closed sets.
an arbitrary measurable set can be well approximated by the
open sets that contain it, and alternatively, by the closed sets it
contains.
Theorem 3.4 Suppose is a measurable subset of . Then, for every :
There exists an open set with and .
There exists a closed set
with and .
If is finite, there
exists a compact set with and .
If is finite, there
exists a finite union of closed cubes such that The notation
stands for the symmetric difference between the sets and , defined by , which
consists of those points that belong to only one of the two sets or .
proof. Part(i) is
just the definition of the measurability.
For the second part, we know that is measurable, so there exists an
open set with and . If
we let , then is closed, , and
. Hence
as
desired.
Comments: If we meet the closed set, we can consider
its complement, because its complement is open set.
For (iii), we first pick a closed set so that and . For each , we let denote the ball centered at the
origin of radius and define
compact sets . Then
is a sequence of measurable
that decreases to , and since
, we conclude that
all large one has .
For (iv), choose a family of closed cubes so that Since ,
the series converges and exists such that . If , then
Invariance properties
of Lebesgue measure
Translation-invariance: If is a measurable set and , then the set is alse
measurable, and .
To prove the measurability of under the assumption that is measurable, we note if is open, , and , then
is open, , and
Relative dilation-invariance: Suppose , and denote by the set . We can then
assert that is measurable
whenever is, and .
Reflection-invariant: Whenever is measurable, so is and .
- algebras and Borel sets
Definition. A - algebra of sets is a
collection of subsets of that is closed under
countable unions, countable intersections, and complements.
NOTE: We should compare the definition of - algebra with the topology. In
the definition of the topology, we only need to guarantee that
finite intersections is closed, but the union
can be arbitrary, instead of countable. This is a difference of
the two important definition.
The collection of all subsets of is of course a - algebra.
Another - algebra, which
plays a vital role in analysis, is the Borel - algebra in , denoted by , which by
definition is the smallest -
algebra that contains all open sets. Elements of this - algebra are called
Borel sets.
We should discuss the existence and uniqueness of the Borel -algebra.
The definition of the Borel -algebra will be meaningful once we
have defined the term “smallest,” and shown that such a -algebra exists and is unique. The
term "smallest" means that if is any -algebra that contains all open
sets , then
necessarily . Since we observe that any intersection
(not necessarily countable) of -algebras is again a -algebra, we may define as the
intersection of all -algebras
that contain the open sets. This shows the existence
and uniqueness of the Borel -algebra.
Since open sets are measurable, we conclude that the Borel -algebra is contained in the -algebra of measurable sets.
Naturally, we may ask if this inclusion is strict.
do there exist Lebesgue measurable sets which are not Borel sets? The
answer is “yes.”
Starting with the open and closed sets, which are the simplest Borel
sets. Next in order would come countable intersections of open
sets; such sets are called sets. Alternatively,
one could consider their complements, the countable union of
closed sets, called the sets.
Corollary 3.5 A subset of is measurable
if and only if differs
from a by a set of
measure zero.
if and only if differs
from an by a set of
measure zero.
proof. Clearly
is measurable whenever it
satisfies either (i) or (ii), since the , , and sets of measure zero are
measurable.
Conversely, if is measurable,
then for each integer we may
select an open set
that contains , and such that
. Then is a that contains , and for all . Therefore for all ; hence has exterior measure zero, and is
therefore measurable.
For the second implication, we simpy apply part (ii) of Theorem 3.4 with
, and take
the union of the resulting closed sets.
Construction of a
non-measurable set
Are all subsets of measurable? In this
section, we answer this question when by constructing a subset of which is not measurable. This
justifies the conclusion that a satisfactory theory of measure cannot
encompass all subsets of
The construction of a non-measurable set uses the axiom of choice, and
rests on a simple equivalence relation among real numbers in .
We write whenever is rational, and note that this is an
equivalence relation since the follwing properties hold:
for every
if , then
if and , then .
Two equivalence classed either are disjoint or coincide, and is the disjoint union of all
equivalence classed, which we write as Now we construct the set by choosing exactly one
element from each , and setting .
Theorem 3.6 The set is not measurable.
proof. The proof
is by contradiction, so we assume that is measurable. Let be an
enumeration of all the rationals in , and consider the translates We claim that the sets are disjoint, and To see why these sets are disjoint, suppose that the
intersection is non-empty. Then there exist rationals
and and with ;
hence Consequently and is rational; hence , which
contradicts the fact that contains only one
representative of each equivalence class.
The second inclusion is straightforward since each is contained in by construction. Finally, if , then for some , and therefore for some . Hence , and the first inclusion holds. Now we may
conclude the proof of the theorem. If were measurable, then so
would be for all , and since the union is
disjoint, the inclusions in (4) yield Since is a
tanslate of , we must
have for all .
Consequently, This is the desired contradiction, since neither nor is possible.
Axiom of choice
That the construction of the set is possible is based on the
following general proposition.
Suppose is a set and is a collection of
non-empty subsets of . (The
indexing set of 's is not
assumed to be countable.) Then there is a function (a "choice
function") such that for all .
In this general form this assertion is known as the axiom of
choice.
The initial realization of the importance of this axiom was in its
use to prove a famous assertion of Cantor, the well-ordering
principle. This proposition (sometimes referred to as
"transfinite induction") can be formulated as follows.
A set is linearly
oredered if there is a binary relation such that:
(a) x for all .
If are distinct, then
either or (but not both).
If and , then .
We say that a set can be
well-orederd if can be linearly oredered in such a way
that every non-empty subset has a smallest element in that ordering (that is, an element
such that for any other ).
EXAMPLE. A simple example of a well-ordered set is
, the positive
integers with their usual ordering. The fact that is well-ordered is an
essential part of the usual (finite) induction principle. More
generally, the well-ordering principle states:
Any set can be
well-ordered.
It is in fact nearly obvious that the well-ordering principle
implies the axiom of choice: if we well-order , we can choose to be the smallest element in
, and in this way we have
constructed the required choice function. It is also true, but not as
easy to show, that the converse implication holds, namely that the axiom
of choice implies the well-ordering principle.
The fact that the axiom of choice and the well-ordering principle are
equivalent is a consequence of the following considerations.
One begins by defining a partial ordering on a set to be a binary relation on the set that satisfies:
x for all .
If and , then .
If and , then .
If in addition or whenever , then is a linear ordering of .
The axiom of choice and the well-ordering principle are then
logically equivalent to the Hausdorff maximal principle:
Every non-empty partially ordered set has a (non-empty) maximal
linearly ordered subset.
In other words, if is patially
ordered by , then contains a non-empty subset which is linearly ordered by and such that if is contained in a set also linearly ordered by , then .
An application of the Hausdorff maximal principle to the collection
of all well-orderings of subsets of E implies the well-ordering
principle for E. However, the proof that the axiom of choice implies the
Hausdorff maximal principle is more complicated.
Measurable functions
With the notion of measurable sets in hand, we now turn our attention
to the objects that lie at the heart of integration theory: measurable
functions.
The starting point is the notion of a characteristic
function of a set , which
is defined by The next step is to pass to the functions that are the
building blocks of integration theory. For the Riemann integral it is in
effect the class of step functions, with each given as
a finite sum where each is a
rectangle, and the are
constants.
However, for the Lebesgue integral we need a more general notion, as
we shall see in the next chapter. A simple function is
a finite sum where each is a
measurable set of finite measure, and the are constants.
Definition and basic
properties
We begin by considering only real-valued functions on , which we allow to take on
the infinite values and
, so that belongs to the extended real numbers
We shall say that is
finite-valued if for all . In the theory that follows, and the
many applications of it, we shall almost always find ourselves in
situations where a function takes on infinite values on at most
a set of measure zero.
A function defined on a
measurable subset of is measurable, if for all
, the set is measurable. To sumplify our notation, we shall often denote
the set
simply by whenever no
confusion is possible.
Definiton. If
is a set, a basis for a topology on is a collection of subsets of (called basis elements) such that
For each , there is at
least one basis element
containing
If , then
there is a basis element
containing such that
If satisfies these
two conditions, then we define the topology ofgenerated by as follows: A subset
of is said to be open in
(that is, to be an element of
) if for each there is a basis such that and
Note: Every basis element is itself an element of
Example 1. Let be the collection of all
circular regions (interiors of circles) in the plane. Then satisfies both conditions for
a basis. The second condition is illustrated in Figure 13 1. In the
topology generated by ,
a subset of the plane is open if
every x in lies in some circular
region contained in
Example 2. Let be the collection of all
rectangular regions (intenors of rectangles) in the plane, where the
rectangles have sides parallel to the coordinate axes Then satisfies both
conditions for a basis. The second condition is illustrated in Figure
13. 2; in this case, the condition is trivial, because the intersection
of any two basis elements is itself a basis element (or empty) As we
shall see later, the basis generates the same
topology on the plane as the basis given in the preceding
example.
Example 3. If is any set, the collection of all
one-point subsets of is a basis
for the discrete topology on
We need to check that the collection is indeed a topology on .
Let us check now that the collection generated by the basis is a topology on .
If is the empty set, it
satisfies the defining condition of openness vacuously. Now let us take
an indexed family of elements of and show that Given , there is
an index such that . Since is open, there is an basis
element such that , then , so that is open, by definition.
Now let us take two elements
and of and show that . Given , choose a basis element
containing such that , and containing such that . Hence , and by the second
condition for a basis, there is a basis element such that . Then , so , by definition.
Finally, we show by induction that any finite intersection of
elements of is in . This fact is trivial for
; we suppose it true for and prove it for . Now By hypothesis, ; by the result just proved
the intersection of and
also belong to .
Thus we have checked that collection of open sets generated by a basis
is, in fact, a
topology.
Another way of describing the topology generated by a basis is given
in the following lemma:
Lemma 13.1. Let be a set,let be a basis for a topology
on . Then equals the collection of all
unions of elements of
proof. we need to
prove any subset of satisfies the equation First
given a collection of elements of ,there are also elements of
. Because is a topology, their union is
in . Conversely, given
. Choose each
, there exists an element
of such that , thus ,
and it is obvious that , Thus, .
This lemma states that every open set in can be expressed as a union of basis
elements. This expression for is
not, however, unique. Thus the use of the term "basis" in
topology differs drastically from its use in linear algebra,
where the equation expressing a given vector as a linear combination of
basis vectors is unique.
We have described in two different ways how to go from a basis to the
topology it generates. Sometimes we need to go in the reverse direction,
from a topology to a basis generating it. Here is one way of
obtaining a basis for a given topology; we shall use it
frequently.
Lemma 13.2. Let be a topological space. Suppose that
is a collection of open
sets of such that for each open
set of and each in , there is an element of such that . Then is a basis for the topology
of .
proof. We must
show that is a basis,
thus we need to check the two conditions of definition of basis, given
, since is itself an open set, there is by
hypothesis an element of such that . To check the second
condition, let ,
where and are elements of . Since and are open, so is , then there exists by
hypohesis an element such that .
Let be a collection of
open sets of ; we must show that
the topology
generated by equals the
topology . First, note
that if and if
, then there is by hypohesis
an element of such that . If follows that , by definition.
Conversely, if , then by the preceding lemma. Since each
element of belongs to
and is a topology, also belong to .
We can use the bases to determine whether one topology is
finer than another.
Lemma 13.3. Let and be bases for the two
topologies and on , respectively.Then the following states
are equivalent:
is finer
than .
For each and each
basis element
containing , there is a basis
element
such that .
proof. (2) (1). Given a element of , we wish to show that . Let , since generate , there is a basis element
such that . Condition (2) tell us
that there exist an element such that , then , so , by definition.
(1) (2). Given a
element and , then , and by condition (1),
we have , thus . Since is generated by , there is an element
such that .
Example 4. We can see that the collection of all circular regions in
the plane generates the same topology as the collection of all rectangular
regions.
We now define three topologies on the real line , all of which are of
interest.
Definition. If is a collection of all open
intervals in the real line, the topology generated by is called the
standard topology on the real line. WhWhenever we
consider , we shall
suppose it is given this topology unless we specifically state
otherwise. If is
the collection of all half-open intervals of the form where , the
topology generated by is called the lower
limit topology on . When
is given the lower limit
topology, we denote it by . Finally, let denote the set of all numbers of the
form , for , and let be the collection
of all open intervals, along
with the set of the form . The topology generated by will be called
- topology on
. When is given by this topology, we
denote it by
It is easy to see that all three of these collections are bases;
in each case, the intersection of two basis elements is either
another basis element or is empty. The relation between these
topologies is the following.
Lemma 13.4. The topologies of and are strictly finer than the
standard topology on ,
but are not comparable with another.
proof. Let be the topologies of ,
respectively. Given a basis element for and a point , there exists a basis element
such
that ,
thus is finer than
by the Lemma
13.3.
A similar argument applies to . Given a basis element for and a point , there exist a basis element
that contains . On the other hand,
given a basis element for and point of , there is no open interval contains
and lies in .
Subbasis
Definition. A subbasis for a topology on is a collection of subsets of whose union equals . The topology generated by the
subbasis is
defined to be the collection of all unions of finite
intersections of elements of .
We must of course check that is a topology. By the
definition, we know that the elements of are all the finite
intersections of elements of , and we define the
collection of all finite intersections of elements of is . By the lemma 13.1, it will
suffice to show that is
a basis, and is a
topology. So we just need to check the two conditions of basis for . Given , it belongs to an element of
and hence to an element
of ; this is the first
condition for a basis. To check the second condition, let be two elements of , then the is also a
finite intersection of elements of , so it belongs to .Obviously, .
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